The Purplemath Forums

by **EdSewersPark** on Tue May 24, 2011 11:05 pm

Question: $\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

I'm guessing I have to get rid of the fractions by multiplying by the LCD, but I feel like I am missing a step in even getting that.

**Sponsor**

by **stapel_eliz** on Wed May 25, 2011 12:53 am

EdSewersPark wrote:Question: $\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

I'm guessing I have to get rid of the fractions by multiplying by the LCD...

Yes, that would be the first step. You may want to factor the lower fraction's denominator first, though, so you're clearly aware of what the LCM actually is. :wink:

by **MrAlgebra** on Wed May 25, 2011 12:17 pm

EdSewersPark wrote:Question: $\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

I'm guessing I have to get rid of the fractions by multiplying by the LCD, but I feel like I am missing a step in even getting that.

I would be inclined to start like this. Keeping in mind that

$\frac{\frac{a}{b}}{\frac{c}{d}} = (\frac{a}{b})(\frac{d}{c})$

We have

$\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

$(5+\frac{x+3}{x-1})(\frac{1-x^2}{3x-1})$

Then find the LCD for $(5+\frac{x+3}{x-1})$ and go from there.

by **EdSewersPark** on Wed May 25, 2011 5:02 pm

$\frac{(x-1)}{(x-1)}$ $\frac{(5)}{(1)}$ and $\frac{(x+3)}{(x-1)}$ $\frac{(1)}{(1)}$ which equals $\frac{5x-5}{x-1}\,+\,\frac{x+3}{x-1}$ which equals $\frac{6x-2}{x-1}$

Then I have $(\frac{6x-2}{x-1})(\frac{1-x^2}{3x-1})$

This is where I draw a blank, do I need to change $(1-x^2)$ so that it appears in the denominator? Or do I need to find the LCD for $(x-1)$ and $(3x-1)$ ?

by **stapel_eliz** on Wed May 25, 2011 9:41 pm

EdSewersPark wrote: $\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}}$

You have two denominators: x - 1 and 1 - x² = (1 - x)(1 + x). With a "minus" sign, the first denominator becomes -(1 - x). Take the "minus" sign through the numerator, and you get a subtraction on top.

Use the method illustrated here and multiply, top and bottom, by (1 - x²). This will immediately simplify the fraction considerably! :wink:

by **EdSewersPark** on Thu May 26, 2011 10:27 pm

$(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

Can I cancel out both (x-1)'s and (3x-1)'s at this point?

Would 2(1-x) be the wrong answer?

by **stapel_eliz** on Thu May 26, 2011 11:31 pm

EdSewersPark wrote: $(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

Can I cancel out both (x-1)'s and (3x-1)'s at this point?

Would 2(1-x) be the wrong answer?

Since 1 - x and x - 1 are not the same, no, they cannot be "cancelled". Instead, try making the change explained earlier. Then see about cancelling. :wink:

by **MrAlgebra** on Wed Jun 01, 2011 11:41 pm

EdSewersPark wrote: $(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

Can I cancel out both (x-1)'s and (3x-1)'s at this point?

Would 2(1-x) be the wrong answer?

You're right. It should go:

$(\frac{2(3x-1)}{x-1})(\frac{(1-x)(x-1)}{3x-1})$

$(\frac{2}{x-1})(\frac{(1-x)(x-1)}{1})$ Canceled out 3x - 1.

$(\frac{2}{1})(\frac{(1-x)(1)}{1})$ Canceled out x - 1.

$2(1-x)$ All that's left.

by **Donut2CoffeeCup** on Thu Jun 30, 2011 2:08 pm

Tidy up the top, ie: make it a single fraction:

$\frac{5+\frac{x+3}{x-1}}{\frac{3x-1}{1-x^2}} = \frac{\frac{5x-5+x+3}{x-1}}{\frac{3x-1}{1-x^2}} = \frac{\frac{6x-2}{x-1}}{\frac{3x-1}{1-x^2}}$

Now use $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$ to get

$\frac{(6x-2)}{(x-1)} \times \frac{1-x^2}{3x-1}$

Factorise $(6x-2)=2(3x-1)$

And note (very carefully) that $1-x^2=(1-x)(1+x)$

If you can't see this - reverse the step to check: $(1-x)(1+x)=1-x^2$

So far we have:

$\frac{2(3x-1)}{(x-1)} \times \frac{(1-x)(1+x)}{3x-1}$

To set up the `cancelling' of the $(x-1)$ term write $(x-1) = -(1-x)$ , so that

$-\frac{2(3x-1)}{(1-x)} \times \frac{(1-x)(1+x)}{3x-1}$

Cancel the $(1-x)$ terms:

$-\frac{2(3x-1)}{1} \times \frac{(1+x)}{3x-1}$

Cancel the $(3x-1)$ terms so that you get:

$-\frac{2}{1} \times \frac{(1+x)}{1} = -2(1+x)$

The Purplemath Forums

Simplifying Complex Fraction

Simplifying Complex Fraction

Re: Simplifying Complex Fraction

Re: Simplifying Complex Fraction

Re: Simplifying Complex Fraction

Re: Simplifying Complex Fraction

Re: Simplifying Complex Fraction